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Magnitude of a vector in polar coordinates

  • #1
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Homework Statement


What is the magnitude of the velocity vector if ##\vec{v} = 4 \hat{r} + 6 \hat{\theta}##

Homework Equations




The Attempt at a Solution


I know how do do this in Cartesian coordinates (use the Pythagorean theorem), but not so sure how to do it in polar coordinates.
 

Answers and Replies

  • #2
jambaugh
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Though the point at which this vector is defined may be given in polar coordinates, the basis expansion [itex]\hat{r},\hat{\theta}[/itex] by which the vector is expressed is still an ortho-normal basis so your vector's magnitude still is the root summed square of components.
 
  • #3
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Though the point at which this vector is defined may be given in polar coordinates, the basis expansion [itex]\hat{r},\hat{\theta}[/itex] by which the vector is expressed is still an ortho-normal basis so your vector's magnitude still is the root summed square of components.
Cool! Now I have a similar question.

I am given that ##\dot{r} = 4 ~m/s## and ##\dot{\theta} = 2## rad/s, and when a particle is 3 m from the origin, I am asked to find the magnitude of the velocity and the acceleration. This is easy enough.

From the givens, we know that ##\vec{v} = 4 \hat{r} + 6 \hat{\theta}##. Then we have a magnitude of ##\sqrt{52}##. Then to find the acceleration, we can do either of two things: ##a = \frac{v^2}{r} = \frac{52}{3} = 17.33## m/s. But also, we could use the given information to derive the acceleration vector, which would be ##\vec{a} = -12 \hat{r} + 16 \hat{\theta}##. We find the magnitude of this is 20 m/s^2, not the same as the 17.33 we got earlier. What is going on?
 
  • #4
ehild
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Cool! Now I have a similar question.

I am given that ##\dot{r} = 4 ~m/s## and ##\dot{\theta} = 2## rad/s, and when a particle is 3 m from the origin, I am asked to find the magnitude of the velocity and the acceleration. This is easy enough.

From the givens, we know that ##\vec{v} = 4 \hat{r} + 6 \hat{\theta}##. Then we have a magnitude of ##\sqrt{52}##. Then to find the acceleration, we can do either of two things: ##a = \frac{v^2}{r} = \frac{52}{3} = 17.33## m/s. But also, we could use the given information to derive the acceleration vector, which would be ##\vec{a} = -12 \hat{r} + 16 \hat{\theta}##. We find the magnitude of this is 20 m/s^2, not the same as the 17.33 we got earlier. What is going on?
The acceleration is v2/r in case of uniform circular motion when the velocity is perpendicular to the radius and the acceleration is centripetal only.
 
  • #5
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The acceleration is v2/r in case of uniform circular motion when the velocity is perpendicular to the radius and the acceleration is centripetal only.
So is the correct answer for the magnitude of the acceleration 17.33 m/s^2?
 
  • #6
ehild
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So is the correct answer for the magnitude of the acceleration 17.33 m/s^2?
No. Is it uniform circular motion?
 
  • #7
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No. Is it uniform circular motion?
Oops, I meant is the correct magnitude for the acceleration 20 m/s^2? (I looked at the wrong number)
 
  • #8
ehild
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Oops, I meant is the correct magnitude for the acceleration 20 m/s^2? (I looked at the wrong number)
That is correct.
 
  • #9
jambaugh
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An important feature of this problem is that the local rectilinear coordinate system defined by [itex]\hat{r},\hat{\theta}[/itex] is moving. Thus in calculating the acceleration vector you must take into account the time rate of change of the basis vectors.

In general: [itex] \hat{r} = \hat{i}\cos(\theta) + \hat{j}\sin(\theta)[/itex] so then [itex]\dot{\hat{r}} = \dot{\theta}(-\hat{i}\sin(\theta)+ \hat{j}\cos(\theta)) \equiv \dot{\theta}\hat{\theta}[/itex]. Similarly [itex]\dot{\hat{\theta}} = - \dot{\theta}\hat{r}[/itex]. With these worked out you can take the time derivatives of the position vector to get velocity and acceleration in the local (normalized) coordinate basis:
Velocity: [itex]\vec{r} = r\hat{r},\quad \vec{v}=\dot{\vec{r}} = \dot{r}\hat{r} + r\dot{\hat{r}}=\dot{r}\hat{r} + r\dot{\theta}\hat{\theta}[/itex]
Acceleration:
[itex]\vec{a}=\dot{\vec{v}}=\ddot{r}\hat{r} + \dot{r}\dot{\hat{r}}+ \dot{r}\dot{\theta}\hat{\theta} + r\ddot{\theta}\hat{\theta} + r\dot{\theta}\dot{\hat{\theta}}= [\ddot{r}-r\dot{\theta}^2]\hat{r} +[ 2 \dot{r}\dot{\theta} + r\ddot{\theta}]\hat{\theta} [/itex]

Knowing your velocity is [itex]4\hat{r} + 6\hat{\theta}[/itex] you know that [itex]\dot{r}=4, r\dot{\theta} = 6[/itex] and given [itex] r = 3meters[/itex] you have [itex]\dot{\theta} = 2[/itex].

The acceleration vector is then:
[itex] [\ddot{r}-(3)(2)^2]\hat{r} +[ 2 (4)(2) + (3)\ddot{\theta}]\hat{\theta} = [\ddot{r}-12]\hat{r}+[16+\ddot{\theta}]\hat{\theta}[/itex]
You have not given enough information to solve the problem. You need the 2nd derivatives of the polar coordinates w.r.t. time. Presumably these are zero??? You haven't stated the full problem as given to you. Was there something stating constant rate of change of theses?
 
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